A child stands at the centre of turntable with his two arms out stretched. The turntable is set rotating with an angular speed of 40 rpm. How much is the angular speed of the child if he folds his back and thereby reduces his moment of inertia to `(2)/(3)` times the initial value ? Assume that the turntable rotates without friction

Explanation

Here `omega_(1) = 40 rpm, I_(2) = 2/5 I_(1)`
By the principle of conservation of angular momentum,
`I_(1)omega_(1) = I_(2)omega_(2)` or `I_(1) xx 40 = 2/5 I_(1)omega_(2)` or `omega_(2) = 100` rpm.
(ii) Initial kinetic energy of rotation.
`2/5 I_(1)omega_(1)^(2) = 2/5 I_(1) (40)^(2) = 800 I_(1)`
Net kinetic energy of rotation
`2/5 I_(2)omega_(2)^(2) = 1/2 xx 2/3 I_(1) (100)^(2) = 2000 I_(1)`
`("New K.E.")/("Initial K.E.") = (2000 I_(1))/(800 I_(1)) = 2.5`
Thus the child's new kinetic energy of rotation is 2.5 times its initial kinetic energy of rotation. This increase in kinetic energy is due to the internal energy of the child which he uses in folding his hands back frrom the out stretched position.